Answer
a). $a=-3, b=-9$.
b). $a=-3$, $b=-24$
Work Step by Step
a. Given the function $f(x)=x^3+ax^2+bx$, we have $f'(x)=3x^2+2ax+b$ and $f''(x)=6x+2a$. For the function to have a local maximum at $x=-1$ and a minimum at $x=3$, we have $f'(-1)=f'(3)=0$, which gives
$3(-1)^2+2a(-1)+b=3(3)^2+2a(3)+b=0$
or
$\begin{cases} 2a-b=3\\6a+b=-27 \end{cases}$
which gives $a=-3, b=-9$. Check $f''(-1)=-6-6=-12\lt0$, which confirms the value is a local maximum. Check $f''(3)=18-6=12\gt0$ which confirms the value is a local minimum.
b. For the function to have a point of inflection at $x=1$, let $f''(1)=0$; we have $6(1)+2a=0$ or $a=-3$. A local minimum at $x=4$ leads to $f'(4)=0$ or $3(4)^2+2(-3)(4)+b=0$ which gives $b=-24$
Check $f''(4)=6(4)+2(-3)=18\gt0$, which confirms the value is a local minimum.
Check the signs of $f''$ across $x=1$ at $a=-3$ to get: $..(-)..(1)..(+)..$, which confirms the inflection point.
